Final temperature of the aluminum is 41.8 °C
We have the equation for energy
E = mc∆T
Here m = 55 g = 0.055 kg
∆T=T-27.5
Specific heat capacity of aluminum = 921.096 J/kg.K
E = 725 J
Substituting
E = mc∆T
725 0.055 x 921.096 x (T - 27.5)
T-27.5 14.31
T=41.81° C = 41.8 °C
41.8 °C
Final temperature of the aluminum = 41.8°C
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Answer:
D
Explanation:
I'm so sorry!! I got it wrong sorry!
Answer:
219.78 mL of the stock solution are needed
Explanation:
First, we take a look at the desired Al2(SO4)3 working solution. We are told that the we need 400 mL of an aqueous aluminum sulfate solution 1.0 M. Let's see how many moles of the compound we have in the desired volume:
1000 mL Al2(SO4)3 solution ----- 1 mole of Al2(SO4)3
400 mL Al2(SO4)3 solution ----- x = 0.4 moles of Al2(SO4)3
To reach the desired concentration in the working solution we need 0.4 moles of Al2(SO4)3 in 400 mL, so we calculate the volume of the stock solution needed to prepare the working solution:
1.82 moles Al2(SO4)3/L = 1.82 M → This is the molar concentration of the stock solution.
1.82 moles of Al2(SO4)3 ----- 1000 mL
0.4 moles of Al2(SO4)3 ----- x = 219.78 mL
So, if we take 219.78 mL of the 1.82 M stock solution, we put it in a graduated cylinder and we dilute it to 400 mL, we would obtain a 1.0 M Al2(SO4)3 solution.
A corn oil. Because oil is treated with a solvent. After that it is refined which gets rid of free fatty acids. Finally, it is sent through steam distillation to get rid of volatile organic compounds.
